A 2.0 µC point charge travels a distance of 0.20 cm between two parallel charged plates from the negative end toward the positive end. The electric field has a strength of 500.0 N.C. What is the change in voltage? (µC=1.0*10^-6 C)

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A 2.0 µC point charge travels a distance of 0.20 cm between two parallel charged plates from the negative end toward the positive end. The electric field has a strength of 500.0 N.C. What is the change in voltage? (µC=1.0*10^-6 C)

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A. -100 V B. +1.0 V C. 0 V D. +0.20 mV
the subsequent relationship holds: \[\Delta V = Ed\] |dw:1433273196060:dw|
ok! so what happens next? :/

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Other answers:

500.0 x 0.2
=100
+1.0 V is your answer! :)
ohh okay!! thank you!!:D
no, it is: \[\Large \Delta V = Ed = 500 \times 2 \times {10^{ - 3}} = ...\] since the distance is 0.2 cm = 2*10^(-3) meters
so +1.0V is the solution?
yes! since our charge is approaching to the positive plane, so its potential is increasing, namlely final potential is larger than initial potential so the difference between them is positive
yay! thank you both!!
:)

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